Java-8 addAll Objects

Question

Java 8 中有更好的方法吗？

final List<InstitutionUserConnection> allInstitutionUserConnections = new ArrayList<>();
for (final Institution institution : institutionsOfUser) {
    allInstitutionUserConnections
        .addAll(institutionUserConnectionService.getActiveInstitutionUserConnectionsByInstitution(institution));
}

- quma

Answer 1

是的，这就是flatMap操作的作用：

List<InstitutionUserConnection> allInstitutionUserConnections =
    institutionsOfUser.stream()
                      .flatMap(institution -> institutionUserConnectionService.getActiveInstitutionUserConnectionsByInstitution(institution).stream())
                      .collect(Collectors.toList());

如果该方法抛出已检查异常，则需要捕获并用日志记录进行处理，重新抛出未经检查的异常和/或返回默认值。

List<InstitutionUserConnection> allInstitutionUserConnections =
    institutionsOfUser.stream()
                      .flatMap(institution -> {
                          try {
                              return institutionUserConnectionService.getActiveInstitutionUserConnectionsByInstitution(institution).stream();
                          } catch (TheCheckedThrownException e) {
                              // do something here, like throw an unchecked exception
                          }
                      })
                      .collect(Collectors.toList());

- Tunaki

---

institutionUserConnectionService.getActiveInstitutionUserConnectionsByInstitution(institution)抛出一个异常。还有没有处理的可能性？非常感谢！ - quma

Answer 2

不必将内部列表平铺映射到一个流中，你也可以直接将其映射到List，然后使用自定义收集器来追加元素。使用收集器的优点在于，它消耗更少的性能，因为它使用了1个大的addAll调用，而不像Collectors.toList()那样使用单独的小的add调用。

基于此的解决方案如下：

List<InstitutionUserConnection> result = institutionsOfUser.stream().
        .map(institutionUserConnectionService::getActiveInstitutionUserConnectionsByInstitution)
        .collect(Collector.of(ArrayList::new, List::addAll,
                (left, right) -> { left.addAll(right); return left; }));

- Ferrybig

---

对于“addAll”的引用不明确，既可以匹配“addAll（Collection <? extends E>）”，也可以匹配“addAll（int，Collection <? extends E>）”。 - user1608790